RD Sharma Class 12 Exercise 17.3 Maxima And Minima Solutions Maths

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RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise
Maxima and Minima Excercise: 17.3
Maxima and Minima excercise 17.3 question 1 (iii)
Maxima and Minima Excercise 17.3 question 1 (v)
Maxima and Minima exercise 17.3 question 1 (vi)
Maxima and Minima excercise 17.3 question 1 (ix)Answer:
Maxima and Minima exercise 17.3 question 2 (i)
Maxima and Minima exercise 17.3 question 3
Maxima and Minima excercise 17.3 question 4
Maxima and Minima exercise 17.3 question 8
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RD Sharma Chapter-wise Solutions

RD Sharma Class 12 Solutions Chapter 17 Maxima and Minima - Other Exercise

Maxima and Minima Excercise: 17.3

Maxima and Minima excercise 17.3 question 1 (i)

Answer:

Point of local maxima is at x=1 and it’s local maximum value is 68. Also point of local minima is at x=5,-6 and it’s local minimum values are -316 and -1647.

Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) < 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value
Given:
$f(x)=x^{4}-62 x^{2}+120 x+9$
Explanation:
It is given that
$\begin{aligned} &f(x)=x^{4}-62 x^{2}+120 x+9 \\ &f^{\prime}(x)=4 x^{3}-62 * 2 x+120 \end{aligned}$
To find critical values ,
$\begin{aligned} &f^{\prime}(x)=0 \\ &4 x^{3}-124 x+120=0 \\ &4\left(x^{3}-31 x+30\right)=0 \end{aligned}$
On solving,
critical values of x are 5,1,-6.
Now,
$f^{\prime \prime}(x)=12 x^{2}-1$
At x=5,
$\begin{aligned} f^{\prime \prime}(5) &=12(5)^{2}-124 \\ &=176>0 \end{aligned}$
So, x=5 is point of local minima.
At x=1,
$\begin{aligned} f^{\prime \prime}(1) &=12(1)^{2}-124 \\ &=-112<0 \end{aligned}$
So, x=1 is point of local maxima.
At x=-6,
$\begin{aligned} f^{\prime \prime}(-6) &=12(-6)^{2}-124 \\ &=308>0 \end{aligned}$

So, x=-6 is a point of local minima.
The local max. value at x=1 is
$\begin{aligned} f(1) &=(1)^{4}-62(1)^{2}+120(1)+9 \\ &=68 \end{aligned}$
And local min. value at x=5 & x=-6 are,
$\begin{aligned} f(5)=&(5)^{4}-62(5)^{2}+120(5)+9 \\ &=-316 \\ f(-6) &=(-6)^{4}-62(-6)^{2}+120(-6)+9 \\ &=-1647 \end{aligned}$
Hence, point of local maxima is 1 & its maximum value is 68.
Also, point of local minima are 5&-6 and its minimum values are -316 &-1647 respectively.

Maxima and Minima excercise 17.3 question 1 (ii)
Answer:

point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-6 x^{2}+9 x+15$
Explanation:
We have,

$\begin{aligned} f(x) &=x^{3}-6 x^{2}+9 x+15 \\ f^{\prime}(x) &=3 x^{2}-6.2 x+9 \\ &=3 x^{2}-12 x+9 \\ \therefore f^{\prime}(x) &=3\left(x^{2}-4 x+3\right) \\ \& f^{\prime \prime}(x) &=3(2 x-4) \\ &=6 x-12 \end{aligned}$
To find maxima and minima.
$\begin{aligned} &\qquad f^{\prime}(x)=0 \\ &3\left(x^{2}-4 x+3\right)=0 \\ &\quad x^{2}-4 x+3=0 \\ &X=3 \text { or } x=1 \\ &\text { At } x=3, \\ &f^{\prime \prime}(3)=6(3)-12 \\ &=6>0 \end{aligned}$
At x = 3,
f’’(3) = 6(3)-12
=6>0
So, x = 3 is point of local minima
At x = 1,
f’’(1) = 6(1)-12
=-6<0
So, x = 1 is point of local maxima
Now, local maximum value at x= 1 is
$\begin{aligned} &f(1)=(1)^{3}-6(1)^{2}+9(1)+15 \\ &f(1)=19 \end{aligned}$

And local min. value at x=3 is
$\begin{aligned} &f(3)=(3)^{3}-6(3)^{2}+9(3)+15 \\ &f(3)=15 \end{aligned}$
Thus, point of local maxima is 1 & its max. value is 19 & point of local minima is 3 & its value is 15.

Maxima and Minima excercise 17.3 question 1 (iii)

Answer:
Point of local maxima value is -2 and it’s local maximum value is 0. Also point of local minima is 0and it’s local minimum value is -4
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x-1)(x+2)^{2}$
Explanation:
We have,
$\begin{aligned} f(x) &=(x-1)(x+2)^{2} \\ f^{\prime}(x) &=(x+2)^{2}(1)+(x-1) 2(x+2) \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=(x+2)(x+2+2 x-2) \\ &=(x+2)(3 x) \\ f^{\prime \prime}(x) &=3 x(1)+(x+2) 3 \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &=6 x+6 \end{aligned}$
For max and min, f’(x)=0,
$\begin{aligned} &(x+2)(3 x)=0 \\ &x=0 \& x=-2 \end{aligned}$
At x=0,
$f^{\prime \prime}(0)=6(0)+6=6>0$
So, x= 0 is point of local minima
$f^{\prime \prime}(0)=6(0)+6=6>0$
At x=-2,
$\begin{aligned} f^{\prime \prime}(0) &=6(-2)+6 \\ &=-6<0 \end{aligned}$
So, x = -2 is point of local maxima
So, local max. value at x=-2 is
f(-2)=(-2-1)(-2+2)^{2}=0
And local min. value at x= 0 is
$f(0)=(0-1)(0+2)^{2}=-4$
Thus, point of local maxima is -2 & its max. value is 0 & point of local minima is 0 & its value is -4.

Maxima and Minima exercise 17.3 question 1 (iv)

Answer:

Point of local maxima is 2 and it’s local maximum value is $\frac{1}{2}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f"(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0$
Explanation:
We have,
$\begin{gathered} f(x)=\frac{2}{x}-\frac{2}{x^{2}} \quad, x>0 \\ \therefore f^{\prime}(x)=-\frac{2}{x^{2}}+\frac{4}{x^{3}} \\ f^{\prime \prime}(x)=\frac{4}{x^{2}}-\frac{12}{x^{4}} \end{gathered}$

For maxima and minima, f’(x)=0
$\begin{array}{r} -\frac{2}{x^{2}}+\frac{4}{x^{3}}=0 \\ -\frac{2(x-2)}{x^{3}}=0 \\ x=2 \end{array}$
At x=2,
$\begin{aligned} f^{\prime \prime}(2) &=\frac{4}{(2)^{3}}-\frac{12}{(2)^{2}} \\ &=\frac{4}{8}-\frac{12}{16} \\ &=\frac{1}{2}-\frac{3}{4} \\ &=-\frac{1}{4}<0 \end{aligned}$
So, x=2 is point of local maxima.
So, local max. value at x= 2 is
$\begin{aligned} f(2) &=\frac{2}{2}-\frac{2}{(2)^{2}} \\ &=1-\frac{2}{4}=\frac{1}{2} \end{aligned}$
Hence, point of local maxima is 2 & its max. value is $\frac{1}{2}$

Maxima and Minima Excercise 17.3 question 1 (v)

Answer:
Point of local minima value is x=-1 and it’s local minimum value is $-\frac{1}{e^{.}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f"(c_1) > 0$ then $c_1$ is point of local minima.
If $f"(c_2) < 0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given: $f(x)=x e^{\mathrm{x}}$
Explanation:
We have,
$\begin{aligned} &f(x)=x e^{\mathrm{x}} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \ldots \text { using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime}(x)=e^{\mathrm{x}}(x+1) \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+1)+e^{\mathrm{x}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=e^{\mathrm{x}}(x+2) \end{aligned}$
Now, for maxima & minima, f’(x)=0
$\begin{aligned} e^{\mathrm{x}}(x+1) &=0 \\ x+1 &=0 \\ x &=-1 \end{aligned}$
at x=-1,
$f^{\prime \prime}(-1)=e^{-1}(-1+2)=\frac{1}{e}$
X = -1 is point of local minima.
So, local min. value at x= 1 is
$f(-1)=-1 e^{-1}=-1 / e$
Thus, point of local minima is -1 & its min value is $-\frac{1}{e}$

Maxima and Minima exercise 17.3 question 1 (vi)

Answer:
Point of local minima value is 2 and it’s local minimum value is 2.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $\mathrm{c}_{1}$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$\frac{x}{2}+\frac{2}{x}, \underline{x}>0$
Explanation:
We have,
$\begin{aligned} &f(x)=\frac{x}{2}+\frac{2}{x} \quad, x>0 \\ &f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^{2}} \ \\ &f^{\prime \prime}(x)=\frac{4}{x^{3}} \end{aligned}$
For local maxima or minima, we have $f^{\prime}(x)=0$
$\begin{gathered} \frac{1}{2}-\frac{2}{x^{2}}=0 \\ x^{2}=4 \\ x=2 \text { or } x=-2 \end{gathered}$
since, $x>0$
$x = 2$
Now, at x=2 ,
$f(2)=\frac{2}{2}+\frac{2}{2}=2$
Thus, its min. value at 2 is 2.

Maxima and Minima Excercise 17.3 question 1 (vii)
Answer:

Point of local minima value is $-\frac{7}{4}$ and it’s local minimum value is $-\frac{3}{(4)^{\frac{4}{3}}}$
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=(x+1)(x+2)^{\frac{1}{3}}, x>=-2$
Explanation:
we have ,
$\begin{aligned} &f(x)=(x+1)(x+2)^{-} \\ &f^{\prime}(x)=(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{\left(-\frac{2}{3}\right)} \ldots \text{ using } \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=\frac{2}{3(x+2)^{2 / 3}}-\frac{2}{9}(x+1)(x+2)^{\left(-\frac{5}{3}\right)} \ldots \text{ using }\frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \end{aligned}$
For maxima & minima, f’(x)=0
$\begin{aligned} &(x+2)^{\frac{1}{3}}+\frac{1}{3}(x+1)(x+2)^{-\frac{2}{3}}=0 \\ &\frac{1}{3}(x+1)=-(x+2)^{\frac{1}{3}}(x+2)^{\frac{2}{3}} \\ &\frac{1}{3}(x+1)=-(x+2) \\ &x+1=-3 x-6 \\ &x=-\frac{7}{4} \end{aligned}$
At x= $-\frac{7}{4}$
$\begin{aligned} f^{\prime \prime}\left(-\frac{7}{4}\right) &=\frac{2}{3}\left(-\frac{7}{4}+2\right)^{\frac{-2}{3}}-\frac{2}{9}\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\left(-\frac{5}{3}\right)} \\ &=\frac{2}{3}\left(\frac{1}{4}\right)^{-\frac{2}{3}}+\frac{1}{18}\left(\frac{1}{4}\right)^{-\frac{5}{3}}>0 \end{aligned}$
So , $x=-\frac{7}{4}$ is a point of local minima
Now at $x=-\frac{7}{4}$
$\begin{aligned} f\left(-\frac{7}{4}\right) &=\left(-\frac{7}{4}+1\right)\left(-\frac{7}{4}+2\right)^{\frac{1}{3}} \\ &=\left(-\frac{3}{4}\right)\left(\frac{1}{4}\right)^{\frac{1}{3}} \\ &=-\frac{3}{4^{\frac{4}{3}}} \end{aligned}$
Thus, point of local minima is $-\frac{7}{4}$ & its value is $-\frac{3}{4^{\frac{4}{3}}}$

Maxima and Minima exercise 17.3 question 1 (viii)

Answer:
Point of local maxima value is 4 and it’s local maximum value is 16. Also point of local minima is -4 and it’s local minimum value is -16
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime}\left(c_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime}\left(c_{2}\right)>0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5$
Explanation:
We have,
$\begin{aligned} &f(x)=x \sqrt{32-x^{2}} \quad,-5 \leq x \leq 5 \\ &f^{\prime}(x)=\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}} \ldots u \sin g \frac{d}{d x} u . v=\frac{u d v}{d x}+\frac{v d u}{d x} \\ &f^{\prime \prime}(x)=-\frac{x}{\sqrt{32-x^{2}}}-\left(\frac{2 x \sqrt{32-x^{2}}+\frac{x^{3}}{\sqrt{32-x^{2}}}}{\left(32-x^{2}\right)}\right) \ldots u \sin g \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \end{aligned}$
Now for local minima and local maxima
$\begin{aligned} &f^{\prime}(x)=0 \\ &\sqrt{32-x^{2}}-\frac{x^{2}}{\sqrt{32-x^{2}}}=0 \\ &32-x^{2}=x^{2} \\ &x^{2}=16 \\ &x=\pm 4 \end{aligned}$
At x=4,
$\begin{aligned} f^{\prime \prime}(4) &=\frac{-4}{\sqrt{32-4^{2}}}-\left[\frac{\left(8\left(32-4^{2}\right)+4^{3}\right)}{\left(32-4^{2}\right)\left(\sqrt{32-4^{2}}\right.}\right] \\ &=-1-\frac{192}{64} \\ &=-3<0 \end{aligned}$
So, x=4 is a point of local maxima.
At x=4,
$\begin{aligned} f(4) &=4 \sqrt{32-(4)^{2}} \\ &=4 \sqrt{32-16} \\ &=16 \end{aligned}$
At x=-4,
$\begin{aligned} f^{\prime \prime}(-4) &=-\frac{4(-4)}{\sqrt{32-(-4)^{2}}}-\left[\frac{8\left(32-4^{2}\right)-(-4)^{3}}{\left(32-4^{2}\right) \sqrt{32-4^{2}}}\right] \\ &=0+2=3>0 \end{aligned}$
So, x=-4 is the point of local minimum.
At x=-4
$\begin{aligned} f(x) &=(-4) \sqrt{32-(-4)^{2}} \\ &=-4 \sqrt{32-16}=-16 \end{aligned}$
Thus, point of local maxima is at x=4 is and its max. value is 16 and & point of local minima is at x=-4 is and its minimum value is -16.

Maxima and Minima excercise 17.3 question 1 (ix)
Answer:

Point of local maxima value is $\frac{a}{3}$ and it’s local maximum value is $\frac{4 a^{3}}{27}.$ Also point of local minima is a and it’s local minimum value is 0.
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f^{\prime \prime}\left(\mathrm{c}_{1}\right)>0$ then $c_1$ is point of local minima.
If $f^{\prime \prime}\left(\mathrm{c}_{2}\right)<0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x^{3}-2 a x^{2}+a^{2} x, a>0$
Explanation:
We have,
$\begin{aligned} &f(x)=x^{3}-2 a x^{2}+a^{2} x \\ &f^{\prime}(x)=3 x^{2}-4 a x+a^{2} \\ &f^{\prime \prime}(x)=6 x-4 a \end{aligned}$
For maxima and minima f’(x) =0
$\begin{aligned} &3 x^{2}-4 a x+a^{2}=0 \\ &x=\frac{-(-4 a) \pm \sqrt{(-4 a)^{2}-4 \times 3 \times a^{2}}}{2 \times 3} \\ &\quad=\frac{4 a \pm \sqrt{16 a^{2}-12 a^{2}}}{6} \\ &=\frac{4 a \pm 2 a}{6} \\ &x=a \text { or } x=\frac{a}{3} \end{aligned}$
At x= a,
$\begin{aligned} f^{\prime \prime}(a) &=6(a)-4 a \\ &=2 a>0 \end{aligned}$
So, x=a is point of local minima
Now at x=a,
$\begin{aligned} f(a) &=a^{3}-2 a(a)^{2}+a^{2}(a) \\ &=a^{3}-2 a^{3}+a^{3}=0 \end{aligned}$
Also, at $\mathrm{x}=\frac{a}{3}$
$\begin{aligned} f\left(\frac{a}{3}\right) &=\left(\frac{a}{3}\right)^{2}-2 a\left(\frac{a}{3}\right)+a^{2}\left(\frac{a}{3}\right) \\ &=\frac{4 a^{3}}{27} \end{aligned}$
Thus, point of local maxima is $\frac{a}{3}$ & its max. value is $\frac{4 a^{3}}{27}$ & point of local minima is a & its value is 0.

Maxima and Minima excercise 17.3 question 1 (x)

Answer:
Point of local minima value is a and it’s local minimum value is 2a. Also point of local maxima is -a and it’s local maximum value is -2a
Hint:
First find critical values of f(x) by solving f'(x) =0 then find f''(x).
If $f''(c_1) >0$ then $c_1$ is point of local minima.
If $f''(c_2) <0$ then $c_2$ is point of local maxima .
where $c_1$ & $c_2$ are critical points.
Put $c_1$ and $c_2$ in f(x) to get minimum value & maximum value.
Given:
$f(x)=x+\frac{a^{2}}{x}, a>0$
Explanation:
We Have,
$\begin{aligned} &f(x)=x+\frac{a^{2}}{x} \\ &f^{\prime}(x)=1-\frac{a^{2}}{x^{2}} \\ &f^{\prime \prime}(x)=\frac{2 a^{2}}{x^{3}} \end{aligned}$
For maxima and minima ,f’(x)=0
$\begin{aligned} 1-\frac{a^{2}}{x^{2}} &=0 \\ x &=\pm a \end{aligned}$

Maxima and minima exercise 17.3 question 1 (xi)

Answer:

x=1 is a point of local maxima and its maximum value is 1 and x=-1 is a point of local minima and its minimum value is -1.
Hint:
Putting x=1,-1in f''(x)
Given:
$\begin{aligned} &f(x)=x \sqrt{2-x^{2}} \\ &-\sqrt{2} \leq x \leq \sqrt{2} \end{aligned}$
Explanation:
Differentiating f(x) with respect to x
$f^{\prime}(x)=\frac{d}{d x}\left(x \sqrt{2-x^{2}}\right)$
$\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{g}(\mathrm{x})\}+\mathrm{g}(\mathrm{x}) \frac{\mathrm{d}}{\mathrm{dx}}\{\mathrm{f}(\mathrm{x})\}\\ &\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}+\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\mathrm{-}(1) \end{aligned}$
$\begin{aligned} &\text { Using, }\\ &\frac{\mathrm{df}(\mathrm{x})^{\mathrm{n}}}{\mathrm{dx}}=\mathrm{nf}(\mathrm{x})^{\mathrm{n}-1} \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{f}(\mathrm{x}) \text { and }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \pm \mathrm{g}(\mathrm{x})]=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{f}(\mathrm{x})) \pm \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{g}(\mathrm{x}))\\ &\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(2-\mathrm{x}^{2}\right)^{1 / 2}\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2-1} \frac{d}{d x}\left(2-x^{2}\right)\\ &=\frac{1}{2}\left(2-x^{2}\right)^{1 / 2}(-2 x)\\ &=-\frac{x}{\sqrt{2-x^{2}}} \end{aligned}$
$\begin{aligned} &\therefore \operatorname{From}(1), \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}\left(-\frac{\mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}\right)+\sqrt{2-\mathrm{x}^{2} \times 1} \\ &=\frac{-\mathrm{x}^{2}+2-\mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \\ &=\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}} \end{aligned}$
$\begin{aligned} &\text { Put, }\\ &\mathrm{f}^{\prime}(\mathrm{x})=0\\ &\frac{2-2 x^{2}}{\sqrt{2-x^{2}}}=0\\ &\Rightarrow 2-2 x^{2}=0\\ &\Rightarrow 2 \mathrm{x}^{2}=2\\ &\Rightarrow \mathrm{x}^{2}=1\\ &\Rightarrow x=\pm 1\\ &\text { if } x \neq \sqrt{2} \text { and } x \neq-\sqrt{2} \end{aligned}$
These x=1 and x=-1 we the possible point of local maxima and minima
Differentiating f’(x) with respect to x
$\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{2-2 \mathrm{x}^{2}}{\sqrt{2-\mathrm{x}^{2}}}\right) \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(2-2 \mathrm{x}^{2}\right)-\left(2-2 \mathrm{x}^{2}\right) \frac{\mathrm{d}}{\mathrm{dx}} \sqrt{2-\mathrm{x}^{2}}}{\left(2-\mathrm{x}^{2}\right)} \end{aligned}$
Using the formula,
$\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}-\mathrm{u} \frac{\mathrm{d}}{\mathrm{ax}} \mathrm{v}}{\mathrm{v}^{2}} \\ &=\frac{\sqrt{2-\mathrm{x}^{2}} \times(-4 \mathrm{x})-\frac{\left(2-2 \mathrm{x}^{2}\right) 1}{2}\left(2-\mathrm{x}^{2}\right)^{-1 / 2}(-2 \mathrm{x})}{\sqrt{2-\mathrm{x}^{2}} 2} \quad \text { (Applying chain rule) } \\ &=\frac{-4 \mathrm{x}}{\sqrt{2-\mathrm{x}^{2}}}+\frac{\mathrm{x}\left(2-2 \mathrm{x}^{2}\right)}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-4 \mathrm{x}\left(2-\mathrm{x}^{2}\right)+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \\ &=\frac{-8 \mathrm{x}+4 \mathrm{x}^{3}+2 \mathrm{x}-2 \mathrm{x}^{3}}{\left(2-\mathrm{x}^{2}\right)^{3 / 2}} \end{aligned}$
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=1 \\ &\mathrm{f}^{\prime \prime}(1)=\frac{-6(1)+2(1)^{3}}{\left(2-(1)^{2}\right)^{3 / 2}} \\ &=\frac{-6+2}{(1)^{3 / 2}}=-4<0 \end{aligned}$

So x=1 is local maxima and its maximum value f(1)=1

$\begin{aligned} &f^{\prime \prime}(x) \text { at } x=-1 \\ &f^{\prime \prime}(-1)=\frac{-6(-1)+(2)(-1)^{3}}{\left(2-(-1)^{2}\right)^{3 / 2}} \\ &=\frac{6-2}{(\sqrt{1})^{3}}=4>0 \end{aligned}$

So , x = 1 is a point of local maxima and its minimum value f(-1)=-1

Maxima and Minima Excercise 17.3 question 1 (xii)

Answer:
x= $\frac{3}{4}$ is a point of local maxima and its maximum value is 5/4.
Hint:
Using chain rule of Differentiation
Given:
$\begin{aligned} &\mathrm{f}(\mathrm{x})=\mathrm{x}+\sqrt{1-\mathrm{x}} \\ &\mathrm{x} \leq 1 \end{aligned}$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+\sqrt{1-\mathrm{x}}) \\ &=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{1 / 2} \\ &=1+\frac{1}{2}(1-\mathrm{x})^{\frac{1}{2}-1}(-1) \end{aligned}$ [ Using Chain Rule Diffrentiation ]
$\begin{aligned} &=1-\frac{1}{2 \sqrt{1-x}} \\ &=\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}} \\ &\text { Put } f^{\prime}(x)=0 \\ &\frac{2 \sqrt{1-x}-1}{2 \sqrt{1-x}}=0 \\ &\Rightarrow 2 \sqrt{1-x}-\mid 1=0 \text { if } x \neq 1 \\ &\Rightarrow \sqrt{1-x}=\frac{1}{2} \end{aligned}$ [Squaring Both Sides]
$\begin{aligned} &\text { i. } e 1-x=\frac{1}{4} \\ &x=\frac{3}{4} \end{aligned}$
Thus , $x=\frac{3}{4}$ is the point of local maxima and minima
Differentiating f’ with respect to x
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(1-\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{2 \sqrt{1-\mathrm{x}}}\right)\\ &=0-\frac{1}{2} \frac{\mathrm{d}}{\mathrm{dx}}(1-\mathrm{x})^{-1 / 2}\\ &=\frac{1}{2} \times-\frac{1}{2}(1-x)^{-3 / 2}(-1)\\ &\text { (Using chain rule) }\\ &=-\frac{1}{4(1-x)^{\frac{3}{2}}} \end{aligned}$ $\begin{aligned} &\text { Put } \mathrm{f}^{\prime \prime}(\mathrm{x}) \text { at } \mathrm{x}=\frac{3}{4} \\ &\mathrm{f}^{\prime \prime}\left(\frac{3}{4}\right)=\frac{-1}{4\left(1-\frac{3}{4}\right)^{3 / 2}} \\ &=\frac{-1}{4 \times \frac{1}{2 \sqrt{2}}} \\ &=-\frac{\sqrt{2}}{2}<0 \end{aligned}$

So $x = \frac{3}{4}$ is a point of local maxima
The local maximum value at f(x)
$\begin{aligned} &\mathrm{f}\left(\frac{1}{2}\right)=\frac{3}{4}+\sqrt{1-\frac{3}{4}} \\ &=\frac{3}{4}+\frac{1}{2}=\frac{3+2}{4}=\frac{5}{4} \end{aligned}$

Maxima and Minima exercise 17.3 question 2 (i)

Answer:

$x=2$ is the local minima and the local minimum value of $f(x)$ at $x=2$ is 0.
$x=\frac{4}{3}$ is the local maxima and the local maximum value of $f(x)$ at $x=\frac{4}{3}$ is $x=\frac{4}{27}$ .

Hint:
Using chain rule
Given:
$f(x)=(x-1)(x-2)^{2}$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\mathrm{x}-1)(\mathrm{x}-2)^{2}\right) \\ &=(\mathrm{x}-1) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-2)^{2}+(\mathrm{x}-2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1) \\ &=2(\mathrm{x}-1)(\mathrm{x}-2)+(\mathrm{x}-2)^{2} \quad \text { [Using chain rule] } \\ &=2\left(\mathrm{x}^{2}-2 \mathrm{x}-\mathrm{x}+2\right)+\left(\mathrm{x}^{2}-4 \mathrm{x}+4\right) \\ &=2 \mathrm{x}^{2}-4 \mathrm{x}-2 \mathrm{x}+4+\mathrm{x}^{2}-4 \mathrm{x}+4 \\ &=3 \mathrm{x}^{2}-10 \mathrm{x}+8 \end{aligned}$
$\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &3 x^{2}-10 x+8=0 \\ &x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{(10)^{2}-4 \cdot 3 \cdot 8}}{2 \cdot(3)} \\ &=\frac{+10 \pm \sqrt{100-96}}{6} \\ &=\frac{-10 \pm 2}{6} \end{aligned}$
$\begin{aligned} &\mathrm{x}=\frac{+10+2}{6}\\ &=2\\ &\text { And }\\ &x=\frac{+10-2}{6}\\ &=\frac{4}{3} \end{aligned}$

Differentiating $f(x)$ with respect to $x$
$\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-10 \mathrm{x}+\mathrm{i}\right) \\ &=6 \mathrm{x}-10 \end{aligned}$
When put x = 2 in f’’(x)
$\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}$
So x=2 is the local Minima
The Local Minimum value of f(x) at x =2 is 0
Put $x=\frac{4}{3} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})$
$\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}$
The local maximum value of $f(x)$ at $x=\frac{4}{3}$ is $x=\frac{4}{27}$ .

Maxima and Minima excercise 17.3 question 2 (iii)

Answer:
x=1 is point of inflexion
x=-1 is the point of local minimum & $x=\frac{1}{5}$ is the point of local maximum
Hint:
Using chain rule of derivative
Given:
$f(x)=-(x-1)^{3}(x+1)^{2}$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+1)^{2}+(\mathrm{x}+1)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^{3}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \cdot 2(\mathrm{x}+1)+3(\mathrm{x}+1)^{2}(\mathrm{x}-1)^{2}\right] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)[2 \mathrm{x}-2+3 \mathrm{x}+3] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)(5 \mathrm{x}+1) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+\mathrm{x}+5 \mathrm{x}+1\right) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}+6 \mathrm{x}^{3}+\mathrm{x}^{2}-10 \mathrm{x}^{3}-12 \mathrm{x}^{2}-2 \mathrm{x}+5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}-4 \mathrm{x}^{3}-6 \mathrm{x}^{2}+4 \mathrm{x}+1\right) \end{aligned}$
$\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &-1(x-1)^{2}(x+1)(5 x+1)=0 \\ &(x-1)^{2}(x+1)(5 x+1)=6 \\ &\Rightarrow(x-1)^{2}=0 \\ &x-1=0 \\ &x=1 \\ &x+1=0, \quad 5 x+1=0 \\ &x=-1, \quad x=-\frac{1}{5} \end{aligned}$
Thus $x=1$ and $x=-1$ and $x=-\frac{1}{5}$ are the possible points of local minima and maxima
Differentiating f’(x) with respect to x $\begin{aligned} &f^{\prime \prime}(x)=\frac{d}{d x}\left[-\left(5 x^{4}-4 x^{3}-6 x^{2}+4 x+1\right)\right] \\ &=-\left(20 x^{3}-12 x^{2}-12 x+4\right) \\ &=-20 x^{3}+12 x^{2}+12 x-4 \\ &\text { when } x=1 \\ &\begin{aligned} f^{\prime \prime}(1) &=-20(1)^{3}+12(1)^{2}+12(1)-4 \\ &=-20+12+12-4=0 \end{aligned} \end{aligned}$

Thus this test is fail as x=1 is point of inflexion
when $x=-1$
$\begin{aligned} f^{\prime \prime}(-1) &=-20(-1)^{3}+12(-1)^{2}+12(-1)-4 \\ &=20+12-12-4=16>0 \end{aligned}$

So , x = -1 is a point of local minimum

When $x=-\frac{1}{5}$

$\begin{aligned} f\left(-\frac{1}{5}\right)=&-20\left(-\frac{1}{5}\right)^{3}+12\left(-\frac{1}{5}\right)^{2}+12\left(-\frac{1}{5}\right)-4 \\ &=-336 / 125<0 \end{aligned}$
So, x=-1/5 is the point of local maximum

Maxima and Minima exercise 17.3 question 3

Answer:
$a = -\frac{2}{3}$
$\mathrm{b}=-\frac{1}{6}$
Hint:
Put? $\frac{d y}{d x}=0 \text { at } x=1,2$
Given:
$y=\operatorname{alog} x+b x^{2}+x$
Explanation:
Differentiating y with respect to x
$\begin{aligned} &\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{~b} \mathrm{x}+1 \\ &\quad=\frac{\mathrm{a}}{\mathrm{x}}+2 \mathrm{bx}+1 \\ &\text { At } \mathrm{x}=1 \mid \\ &\begin{aligned} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) &=\frac{\mathrm{a}}{1}+2 \mathrm{~b}(1)+1 \\ &=\mathrm{a}+2 \mathrm{~b}+1-(1) \end{aligned} \end{aligned}$
$\begin{aligned} &\text { At } x=2 \\ &\frac{d y}{d x}=\frac{a}{2}+2(b) \cdot 2+1 \\ &=\frac{a}{2}+4 b+1-(2) \end{aligned}$
Since x=1 and x=2 are the extreme values,f’(x)=0 at x=1 & 2
Then equation (1) and (2) become
$\begin{aligned} &a+2 b+1=0-(3)\\ &\frac{a}{2}+4 b+1=0\\ &a+8 b+2=0 \quad-(4)\\ &\text { [Multiplying both sides by 2] } \end{aligned}$
Solving , (3) and (4) , we get
$\begin{array}{r} 6 b-1=0 \\ b=-\frac{1}{6} \end{array}$
Putting $B = -1/6 in (3) , we get$
$\begin{aligned} &a-\frac{1}{3}+120 \\ &a=-\frac{2}{3} \end{aligned}$

Maxima and Minima excercise 17.3 question 4

Answer:
We need to prove that x = e is local maxima
Given:
The given function $\frac{\log x}{x}$
Explanation:
$\text { Let } y=\frac{\log x}{x}$
Then
$\begin{aligned} \frac{d y}{d x} &=\frac{x \cdot \frac{d}{d x}(\log x)-\log (x) \frac{d}{d x}(x)}{x^{2}}, \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{x \times \frac{1}{x}-\log x}{x^{2}} \\ &=\frac{1-\log x}{x^{2}} \end{aligned}$
Put,
$\begin{aligned} &\frac{d y}{d x}=0 \\ &\frac{1-\log x}{x^{2}}=0 \\ &\Rightarrow 1-\log x=0, x \neq 0 \\ &\Rightarrow \log x=1 \\ &x=e^{1}=e \end{aligned}$
Differentiating $\frac{d y}{d x}$ with respect to $x$
$\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}} &=\frac{\mathrm{x}^{2} \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}(1-\log \mathrm{x})-(1-\log \mathrm{x}) \frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{x}^{2}\right)}{\mathrm{x}^{4}} . \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{\mathrm{x}^{2}\left(-\frac{1}{\mathrm{x}}\right)-2 \mathrm{x}(1-\log \mathrm{x})}{\mathrm{x}^{4}} \\ &=\frac{-\mathrm{x}-2 \mathrm{x}+2 \mathrm{xlog} \mathrm{x}}{\mathrm{x}^{4}} \\ &=\frac{-3 \mathrm{x}-2 \mathrm{x} \log \mathrm{x}}{\mathrm{x}^{4}} \end{aligned}$
At x=e,
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\frac{-3 e+2 e \log e}{e^{4}} \\ &=\frac{-3 e+2 e(1)}{e^{4}} \ldots \log e=1 \\ &=-\frac{e}{e^{4}}=-\frac{1}{e^{3}}<0 \end{aligned}$
So x = e is a point of local maxima

Maxima and Minima exercise 17.3 question 5

Answer:
x = 0 is local minima and its minimum value is 2 and x =1 is local maxima and its maximum value is -6
Given:
$f(x)=\frac{4}{x+2}+x$
Explanation:
Differentiating f with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(4(\mathrm{x}+2)^{-1}+\mathrm{x}\right) \\ &=-4(\mathrm{x}+2)^{-2}+1 \\ &=-\frac{4}{(x+2)^{2}}+1 \\ &=\frac{-4+\mathrm{x}^{2}+4 \mathrm{x}+4}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}} \\ &=\frac{\mathrm{x}(\mathrm{x}+4)}{(\mathrm{x}+2)^{2}} \end{aligned}$
$\begin{aligned} &\text { Put } f(x)=0 \\ &\frac{x(x+4)}{(x+2)^{2}}=0 \\ &x(x+4)=0 \\ &\text { i.e } x=0, \text { or } \\ &x=-4 \end{aligned}$
Thus x = 0 and x = -4 are the points of local maxima and minima
Differentiating f’ with respect to x
$\begin{aligned} &\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}^{2}+4 \mathrm{x}}{(\mathrm{x}+2)^{2}}\right) \\ &=\frac{(\mathrm{x}+2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+4 \mathrm{x}\right)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+2)^{2}}{(\mathrm{x}+2)^{4}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{\frac{v d u}{d x}+\frac{u d v}{d x}}{v^{2}} \\ &=\frac{(\mathrm{x}+2)^{2}(2 \mathrm{x}+4)-\left(\mathrm{x}^{2}+4 \mathrm{x}\right) \cdot 2(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left[(\mathrm{x}+2)^{2}-\left(\mathrm{x}^{2}+4 \mathrm{x}\right)\right]}{(\mathrm{x}+2)^{4}} \\ &=\frac{2(\mathrm{x}+2)\left(\mathrm{x}^{2}+4 \mathrm{x}+4-\mathrm{x}^{2}-4 \mathrm{x}\right)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8(\mathrm{x}+2)}{(\mathrm{x}+2)^{4}} \\ &=\frac{8}{(\mathrm{x}+2)^{3}} \end{aligned}$
Put x = 0 in f ‘’ (x)
$\begin{aligned} \mathrm{f}^{\prime \prime}(0) &=\frac{8}{(0+2)^{3}} \\ &=\frac{8}{8} \\ &=1>0 \end{aligned}$
S,x=0is local minima. Hence,its minimum value will be f(0)=2
And x= -4 in f ‘’(x)
$\begin{aligned} &f^{\prime \prime}(-4)=\frac{8}{(-4+2)^{3}} \\ &=\frac{8}{-8} \\ &=-1<0 \end{aligned}$
So x=-4 is local maxima. Hence, its maximum value will be f(-4)=-6

Maxima and Minima exercise 17.3 question 6

Answer:
$x=\frac{\pi}{4}$ is a point of local minima and its local minimum value will be $f\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
$x=\frac{3\pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3\pi}{4}\right)=-1-\frac{3\pi}{2}$
Hint:
$\text { Put } f^{\prime} x=0$
Given:
$f(x)=\tan x-2 x$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} &f(x)=\sec ^{2} x-2\\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0\\ &\sec ^{2} x-2=0\\ &\Rightarrow \sec ^{2} \mathrm{x}=2\\ &\Rightarrow \sec \mathrm{x}=\pm \sqrt{2}\\ &=\sec x=\sqrt{2}\\ &\Rightarrow \mathrm{x}=\frac{\pi}{4}\\ &\text { or, } \sec x=-\sqrt{2}\\ &\text { or, } \sec \mathrm{x}=\pi-\frac{\pi}{4}\\ &=\frac{3 \pi}{4} \end{aligned}$
Thus $x=\frac{\pi}{4}$ or $x=\frac{3\pi}{4}$ is possible points of local maxima and minima
Differentiating f’(x) with respect to x
$\begin{aligned} &\mathrm{f}^{\mathrm{u}}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sec ^{2} \mathrm{x}-2\right] \\ &=2 \sec \mathrm{x}(\sec \mathrm{xtan} \mathrm{x}) \\ &=2 \sec ^{2} \mathrm{x} \tan \mathrm{x} \\ &\text { Put } \mathrm{x}=\frac{\pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x}) \\ &\begin{aligned} \mathrm{f}=\left(\frac{\pi}{4}\right) &=2 \sec ^{2}\left(\frac{\pi}{4}\right) \tan \left(\frac{\pi}{4}\right) \\ &=2(2)(1) \\ &=4>0 \end{aligned} \end{aligned}$
So ,
$\mathrm{x}=\frac{\pi}{4}$ is a point of Local Minima and its local maximum value will be $\mathrm{f}\left(\frac{\pi}{4}\right)=1-\frac{\pi}{2}$
$\begin{aligned} &\text { And Put, }\\ &\mathrm{x}=\frac{3 \pi}{4} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})\\ &\mathrm{f}^{\prime \prime}\left(\frac{3 \pi}{4}\right)=2 \sec ^{2}\left(\frac{3 \pi}{4}\right) \tan \left(\frac{3 \pi}{4}\right)\\ &=2(-\sqrt{2})^{2} \quad(-1)\\ &=-4<0 \end{aligned}$
So $x=\frac{3 \pi}{4}$ is a point local maxima and its local maximum value will be $f\left(\frac{3 \pi}{4}\right)=-1-\frac{3 \pi}{2}$

Maxima and Minima exercise 17.3 question 7
Answer:
$\begin{aligned} &a=3 \\ &b=-9 \& c \in R \end{aligned}$
Hint:
Put,
$\begin{aligned} &\mathrm{f}^{\prime}(-1)=0 \\ &\mathrm{f}^{\prime}(3)=0 \end{aligned}$
Given:
$f(x)=x^{3}+a x^{2}+b x+C$
Explanation:
Differentiating f(x) with respect to x
$\begin{aligned} f^{\prime}(x) &=3 x^{2}+2 a x+b+0 \\ &=3 x^{2}+2 a x+b \end{aligned}$
$\begin{aligned} &\text { Calculate } \mathrm{f}^{\prime}(\mathrm{x}) \text { at } \mathrm{x}=-1 \text { and } \mathrm{x}=3\\ &\begin{aligned} \mathrm{f}^{\prime}(-1) &=3(-1)^{2}+2 \mathrm{a}(-1)=\mathrm{b} \\ &=3-2 \mathrm{a}+\mathrm{b} \\ \mathrm{f}^{\prime}(3) &=3(3)^{2}+2 \mathrm{a}(3)+\mathrm{b} \\ &=27+6 \mathrm{a}+\mathrm{b} \end{aligned} \end{aligned}$
Put and $f^{\prime}(-1) =0$ and $f^{\prime}(3)=0$ as they are the extremum values hence $f^{\prime}(x)=0$
$\begin{aligned} &\therefore 3-2 \mathrm{a}+\mathrm{b}=0 \\ &7-2 \mathrm{a}+\mathrm{b}=-3-(1) \\ &27-6 \mathrm{a}+\mathrm{b}=20 \mid \\ &6 \mathrm{a}+\mathrm{b}=-27-(2) \\ &(1)-(2), \mathrm{wc} \text { get } \\ &-2 \mathrm{a}-6 \mathrm{a}=-3+27 \\ &-8 \mathrm{a}=2 \mathrm{y} \\ &\mathrm{a}=3 \\ &\text { Put } \mathrm{a}=3 \text { in }(1) \\ &-2(3)^{2}-\mathrm{b}=-3 \\ &\mathrm{~b}=-9 \end{aligned}$

Maxima and Minima exercise 17.3 question 8

Answer:
Hence proved.
Hint: $\mathrm{f}^{\prime}(\mathrm{x})=0$
Put,
Given:
$\mathrm{f}(\mathrm{x})=\sin \mathrm{x}+\sqrt{3} \cos \mathrm{x}$
Explanation:
Differentiating fx with respect to x
$\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\cos x-\sqrt{3} \sin \mathrm{x} \\ &\text { Put } \mathrm{f}^{\prime}(\mathrm{x})=0 \\ &\cos \mathrm{x}-\sqrt{3} \sin \mathrm{x}=0 \\ &\Rightarrow \sqrt{3} \sin \mathrm{x}=\cos \mathrm{x} \\ &\Rightarrow \tan \mathrm{x}=\frac{1}{\sqrt{3}} \end{aligned}$
$\Rightarrow x=\frac{\pi}{6}$
Differentiating f’(x) with respect to x ,
$\begin{aligned} &f^{\prime \prime}(x)=-\sin x-\sqrt{3} \cos x \\ &\text { Putt } x=\frac{\pi}{6} \text { at } f^{\prime \prime}(x) \\ &f^{\prime \prime}\left(\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}-\sqrt{3} \cos \frac{\pi}{6} \\ &=-\frac{1}{2}-\sqrt{3} \times \frac{\sqrt{3}}{2} \\ &=-\frac{1}{2}-\frac{3}{2} \\ &=-\frac{4}{2} \\ &=-2<0 \end{aligned}$
So, $x = \frac{\pi}{6}$ is a point local maxima.

Rd Sharma Class 12th exercise 17.3 is an essential part of the RD Sharma solutions for maths. The Class 12 RD Sharma chapter 17 exercise 17.3 solution will contain an important chapter of the NCERT maths book that needs to be studied well. RD Sharma Class 12 Solutions Maxima and Minima Ex 17.3 is going to be based on the chapter Maxima and minima.
Rd Sharma class 12-chapter 17 exercises 17.3 answers that are basic and simple. It will help students learn all formulae well and become skilled at the subject in general. RD Sharma class 12 solutions ex 17.3 has around 21 inquiries.

Solutions identified with the Greatest and least upsides of a function in its space
Definition and which means of greatest
Definition and significance of least
Nearby maxima
Questions identified with Nearby minima
Discover Solutions of First subordinate test for neighborhood maxima and minima
Higher-request subordinate test

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